Dependence of resonant frequency on the length of a ruler Why does the resonance frequency of a ruler increase when the length is lower? I have a relationship from experimenting but I want to know why. 
I have a vibration generator hooked up to the bottom of a secured ruler and test its resonance frequency. Why does it need a higher frequency when shorter in height?
 A: The archetypal model for all oscillations is the Mass on a Spring for which the frequency can be written as $$f \propto \sqrt{\frac{k}{m}}$$ where $k$ is stiffness term and $m$ is load or mass. $k$ is a measure of the restoring force which pulls the load back to the equilibrium position while $m$ represents a tendency to maintain the current motion - ie inertia. 
For a cantilever beam the stiffness $k$ is a measure of the force $F$ required to produce a given amount of deflection $x$ - ie $F=kx$. It is related to length by $k \propto 1/L^3$. The shorter the length the greater the force required to produce a given deflection. At the same time, for a fixed cross section the mass of the beam increases in direct proportion to its length : $m \propto L$. Combining these two factors we get $$f \propto \sqrt{\frac{1}{L^4}}=\frac{1}{L^2}$$
This relationship is confirmed by mathematical analysis of the Free Oscillation of a Cantilever Beam.

Your data follows a power law of the form $f \propto 1/L^n$ with $n = 2.28$ to a high degree of accuracy (see trendline equation in graph). This is in approximate agreement with the prediction $n=2$ from above. 
The small discrepancy might be explained by damping in the form of frictional losses of kinetic energy which is transformed into thermal energy within the beam as it flexes. The frequency dependence $f \propto 1/L^2$ derived above is the natural frequency $f_0$ assuming that there is no damping. In the presence of damping (measured by factor $\gamma$ which is related to the fraction of energy lost in each cycle) the oscillation frequency is reduced to $f_1$ such that $$f_1^2=f_0^2-\gamma^2$$ This reduction in frequency is equivalent to an increase in the length of the beam, hence the increase in the exponent $n$.
